Regex to say that string starts with pattern

byMR

December 29, 2022

Is it possible to write a regex to check that a string starts with:

3 digits
either - or +
4 digits
either - or +
2 digits

Examples of strings the pattern should match:

‘123-1234-12’
‘123-1234-‘
‘123-1234’
‘123-‘
‘123’

Examples of strings that the pattern should not match:

‘1231234-12’
‘123-123412’
‘123-1234-12’
‘123–12’

Here’s some Python code I’ve written which accomplishes the same thing. Just wondering though, it is possible to do this with a regular expression?

import re
def match(string):
    for sep1 in ('-', '+'):
        for sep2 in ('-', '+'):
            if re.search(fr'^\d{{3}}$', string):
                return True
            elif re.search(fr'^\d{{3}}\{sep1}$', string):
                return True
            elif re.search(fr'^\d{{3}}\{sep1}\d{{4}}$', string):
                return True
            elif re.search(fr'^\d{{3}}\{sep1}\d{{4}}\{sep2}$', string):
                return True
            elif re.search(fr'^\d{{3}}\{sep1}\d{{4}}\{sep2}\d{{2}}$', string):
                return True
    return False

expected_pass = [
    '123-1234-12',
    '123-1234-',
    '123-1234',
    '123-',
    '123',
]
for string in expected_pass:
    print('expected pass', match(string))

expected_fail = [
    '123-123412',
    '1231234-12',
    '123--12',
]
for string in expected_fail:
    print('expected fail', match(string))

>Solution :

Try with this one:

^\d{3}(?:[\-+](?:\d{4}(?:[\-+](?:\d{2})?)?)?)?$

Regex Explanation:

^: start of string
\d{3}: starts with three digits
(?:[\-+] ... )?: optional - or +
(?:\d{4} ... )?: further optional 4 digits
(?:[\-+] ... )?: even further optional - or +
(?:\d{2} ... )?: even more further optional 2 digits
$: end of string

Check the regex demo here.

Your Python code would become:

import re

pattern = r'^\d{3}(?:[\-+](?:\d{4}(?:[\-+](?:\d{2})?)?)?)?$'

expected_pass = ['123-1234-12', '123-1234-', '123-1234', '123-', '123']
for string in expected_pass:
    string_match = re.match(pattern, string)
    print('expected pass', string_match)

expected_fail = ['123-123412', '1231234-12', '123--12']
for string in expected_fail:
    string_match = re.match(pattern, string)
    print('expected fail', string_match)

Check the python demo here.